# Root Sum Square (RSS) tolerance of parallel resistors

I am trying to figure out the formula for RSS-ing two parallel resistors, but for the life of me can't figure this out.

For series resistors it is easy, just $${R_1 + R_2} \pm \sqrt{ dR_1^2 + dR_2^2}$$

But what is it for two parallel resistors where you use the formula $$R_{\text{tot}} = {R_1 \cdot R_2\over R_1+R_2}$$?

I was thinking you just work out $$\\small R_1 \cdot R_2 \pm \text{tol}_A\$$ and $$\\small R_1+R_2 \pm \text{tol}_B\$$, then just do the division $$R_{\text{tot}} = {R_1 \cdot R_2\over R_1+R_2}$$ and $$\text{tol} = {\text{tol}_A \over \text{tol}_B}$$ but this is not right? Does anyone know how to do this?

• The search term is "propagation of uncertainty or error": geol.lsu.edu/jlorenzo/geophysics/uncertainties/…. Break down each part of the divider equation and apply each rule. You can also just use sums of partial derivatives which is where the rules come from. Commented Dec 14, 2021 at 14:49
• Hi DKNguyen, this is not RSS though. RSS gives a more realistic estimate for the error. This document is talking about Worst Case Analysis (WCA)
– Edba
Commented Dec 14, 2021 at 15:46
• @Edba but this is not right - you didn't say what it should be - what should it be? Commented Dec 14, 2021 at 15:47
• Read it more carefully. Commented Dec 14, 2021 at 15:48
• Convert to conductances and they add. Note that your example values, the 20K adds 25x the uncertainty of the other. In your comment, the 1.414% error is only true for approximately equal resistors; if one R is sufficiently larger (in series) or smaller (in parallel) you can practically ignore the other's tolerance.
– user16324
Commented Dec 14, 2021 at 20:45

Does anyone know how to do this?

If I make a comparison using worst case analysis (with both parallel resistors being very similar in their basic value), it's the same for series resistors too.

So, if you are able to convert a WCA figure to an RSS figure then, the result is the same.

I mention that both resistors should be similar in their basic value because, that will be the most onerous situation to consider.

In series

$$R_{NET} = R_1(1 \pm\Delta)+R_2(1 \pm\Delta)$$ $$= (R_1+R_2)\cdot (1 \pm\Delta)$$

In parallel

$$R_{NET} = \dfrac{R_1(1 \pm\Delta)\cdot R_2(1 \pm\Delta)}{R_1(1 \pm\Delta)+R_2(1 \pm\Delta)}$$ $$= \dfrac{R_1\cdot R_2}{R_1+R_2}\cdot (1 \pm\Delta)$$

• Could anyone tell me why the above has been downvoted. I'd like to fix the answer if there's a mistake but I've probably gone blind to the issue. Commented Dec 17, 2021 at 5:50